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3 years ago

How much force is needed to stop a 50 kg gymnast if he decelerates at 25 m/s^2

1 answer:

3 0

The force can be calculated by multiplying the mass of the gymnast with her acceleration.

Force = 50 kg × 25 m/s2
Force = 1250 N

A force of at least 1250 N can stop the 50-kg gymnast.

I hope I was able to answer your question. Have a good day.

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2. 2. A drawing that shows the outline of an object is called

madreJ [45]

Lineart

Answer: the answer is lineart

4 0

3 years ago

If a car is moving at a constant velocity of 10 m/s, what is its acceleration?

marin [14]

Answer:

If a car is moving at a constant velocity of 10 m/s, there will be no change in the velocity per time.

Acceleration is the rate of change of velocity

Acceleration=(Final velocity-Initial velocity)/Time

Acceleration=(10m/s-10m/s)/Time= 0 m/s²

<h3>★ 0 m/s² is the right answer. </h3>

4 0

3 years ago

The y component of a vector R of magnitude k = Bcm shown in the figure below is Ky = +6 cm. What is the direction of this vector

WARRIOR [948]

Given

The y-component of vector K is

$K_y=6\text{ cm}$

The magnitude of vector K is , K=8 cm

To find

The angle

$\theta$

Explanation

Resolving K along its y-component we have,

$\begin{gathered} K_y=Ksin\theta \\ \Rightarrow6=8sin\theta \\ \Rightarrow sin\theta=\frac{3}{4} \\ \Rightarrow\theta=48.59\text{ }^o \end{gathered}$

Conclusion

The angle made with the x-axis is

$48.59^o$

6 0

1 year ago

A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho

dlinn [17]

Answer:

$x =\frac{-R}{2}$

Explanation:

From the question we are told that mass

Thin layer radius $= 2R$

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state

Considering the center of mass of both bodies

$xcm=\frac{m*x+m*x)}{2m} =x$

$x =\frac{-R}{2}$

Therefore the enclosing layer moves $x =\frac{-R}{2}$

7 0

3 years ago

I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove

Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
69,816,900 km
0. 466 697 AU

 Distance at Perihelion (point in it's orbit that's closest to the sun):
46,001,200 km
0.307 499 AU 

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.

Mercury's Orbital period = 87.9691 Earth days

1/2 (50%) of that is 43.9845 Earth days

The average of the aphelion and perihelion distances is

1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU

This also happens to be 1/2 of the major axis of the elliptical orbit.

3 0

4 years ago